# Physics 122B Electricity and Magnetism

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"Physics 122B Electricity and Magnetism"

```					       Physics 122B
Electricity and Magnetism
Lecture 9 (Knight: 28.1 to 28.5)

Current and Resistance
April 16, 2007

Martin Savage
The Electron Current
experiments” on a simple system. We
have a parallel plate capacitor that
has been charged, e.g. with glass and
plastic rods. Now we connect the
plates with a wire. What happens?

The plates quickly become neutral,
and we say that the capacitor has
been discharged.

Further study shows that while the
discharge is taking place, the wire
gets warm, a light bulb can be made
to glow, and a compass needle can be
deflected. These are indicators of
current flow in the wire.

11/22/2011            Physics 122B - Lecture 9           2
Charge Carriers and Inertia
Does the discharge occur because positive charges are moving to the
negative plate, or because negative charges are moving to the positive plate?

We have asserted that the current in metals is
caused by the flow of negative electrons. The first              +               +               +
direct evidence that this was the case was provided in                   -           -           -
1916 by the Tolman-Stewart experiment, which showed              +               +               +
-
that negative charges “go to the bottom” of an                   -
-
-

accelerated conductor.                                           +               +               +
-
-       -
We model a metallic conductor as a rigid lattice of           +               +               +
-               -
positive charges pervaded by a “sea” of conduction
electrons, 1 per atom, that move freely in the material.
11/22/2011                  Physics 122B - Lecture 9                                           3
The Electron Current
-
In a metallic conductor in electrostatic
equilibrium, the conduction electrons move                    -

around quite rapidly, but there is no net                     -
movement of charge.

This can be changed by “pushing” on the
sea of electrons with an electric field,
thereby causing the entire sea of
electrons to move in a particular direction,
like a gas or liquid flowing through a pipe.            We define the electron
current i as the number of
The net motion, the “drift speed” vd, is       electrons Ne that pass through
superposed on the random thermal motions          a cross section of wire or other
of the individual electrons, and it is very       conductor in a time interval Dt.
slow, typically around 10-4 m/s.                  In other words: Ne = i Dt.

11/22/2011                   Physics 122B - Lecture 9                          4
Current and Drift Velocity

If the electrons have an average drift
speed vd, then on the average in a time                         This table gives n
interval Dt they would travel a distance Dx in the wire,        for various metals.
where Dx = vd Dt. If the wire has cross sectional area
A and there are n electrons per unit volume in the wire,
then the number of electrons moving through the cross
sectional area in time Dt is Ne = n A Dx = n A vd Dt = i Dt .
Therefore,

i  nAvd
11/22/2011                     Physics 122B - Lecture 9                        5
Question 1

Which wire has the largest electron current?

11/22/2011               Physics 122B - Lecture 9           6
Conservation of Current
Question: An electron current iA flows
to the light bulb, passing point A, where
it delivers some energy and makes the
bulb glow. How much electron current iB
then passes point B?
Answer: All of it! iA=iB. Reason: the
electrons don’t have anywhere else to go.
What goes to the bulb must return from
the bulb. The bulb cannot “use up” the
electrons.
Plumber’s Analogy 1: If water flows
into a constant diameter pipe at 2.0 m/s,
it must flow out of the pipe at the same
speed. It cannot “pile up” in the pipe.

This principle is called “Conservation of Current”.
11/22/2011                     Physics 122B - Lecture 9            7
A Puzzle
We discharge a capacitor that
has been given a charge of Q = 16 nC,
using a copper wire that is 2 mm in
diameter and has a length of L = 20
cm. Assume that the electron drift
speed is vd = 10-4 m/s.
How long does it take to discharge
the capacitor? (Note that L/vd =
0.2m/10-4 m/s = 2000 s = 33.3 min.)
Points to consider:
1. The wire is already full of electrons.
2. The wire contains about 5x1022
conduction electrons.
3. Q = 16 nC requires about 1011 electrons.
4. A length L’ of wire that holds 16 nC
of conduction electrons is 4x10-13 m.
5. L’/vd = 4x10-9 s = 4 ns. That is roughly the discharge time.
11/22/2011                 Physics 122B - Lecture 9               8
Creating a Current…
Non-Static Situation
Suppose you want to slide a book
across a table. If you give it a quick push,
it moves but slows due to friction as soon
as you remove your hand, and its kinetic
energy becomes heat. The only way to
make the book move at a constant speed
is to continue pushing it.

The sea of conduction electrons is
similar to the book. If you push the
electrons, you create a current, but they
are not moving in vacuum, and collisions
with other electrons and atoms soon
dissipates their kinetic energy as heat.
The only way to maintain the current of
electrons is to push them, using an
electric field.
An electron current is a non-equilibrium motion in an E field.
11/22/2011                   Physics 122B - Lecture 9                 9
Establishing the
Electric Field in a Wire (1)
The figure shows two metal wires
attached to the plates of a parallel plate
capacitor, with their ends close together but
not touching. The wires are conductors, so
some of the charge from the capacitor
plates spreads out along the wires as surface
charge. E=0 inside all conductors..plates and
wires.
Now we connect the wires. What
happens? The surface electrons can move,
and do so. In ~10-9 s the sea of electrons
shifts slightly, and the surface charges are
rearranged into a non-uniform distribution of
charges, as shown. Surface charges near the
+ and – plates reflect these charges, but
surface charges become near-neutral half-
way along the wire
11/22/2011                  Physics 122B - Lecture 9   10
Establishing the
Electric Field in a Wire (2)

The figure shows the region of the wire near the neutral midpoint. The
surface charge rings become more positive to the left and more negative
to the right.
In Chapter 26, we found that a ring of charge makes an on-axis E field that:
1. Points away from a positive ring and toward a negative ring;
2. Is proportional to the net charge of the ring;
3. Decreases with distance from the ring.
The non-uniform surface charge distribution creates an E field
inside the wire. This pushes the electron current through the
wire
11/22/2011                 Physics 122B - Lecture 9                      11
Example: The Surface Charge
on a Current-Carrying Wire
Two 2.0 mm diameter charged
rings, with charges ±Q, are 2.0 mm
apart.
What value of Q causes the
electric field at the midpoint to be
E=0.010 N/C?

1        zQ
E                               5 103 N/C
4 0  z 2  R 2 3/ 2

4 0  z  R
2

2 3/ 2

Q                              E  1.6 1018 C  10e
z

11/22/2011                           Physics 122B - Lecture 9   12
Turning the Corner

11/22/2011        Physics 122B - Lecture 9   13
Question 2
Which surface charge distribution will
produce the largest electron current?

11/22/2011                Physics 122B - Lecture 9       14
A Model of Conduction (1)
Suppose E  0 : K  1 mv 2  3 kT
2        2

gives v  105 m/s. However, v  0.

Now turn on an E field. The straight-line
trajectories become parabolic, and because of the
curvature, the electrons begin to drift in the
direction opposite E, i.e., “downhill”.
ax=F/m=eE/m so
vx=vix+ axDt = vix+ Dt eE/m
This acceleration increases an electrons
kinetic energy until the next collision, a “friction”
that heats the wire….energy is imparted to the
atoms of the lattice.
11/22/2011                   Physics 122B - Lecture 9                       15
A Model of Conduction (2)
One Collision                             Time-History over many Collisions

___    ___
eE ___ eE ___ e
vd  vx  vix     Dt    Dt  E
m       m     m
ne A              ( Collision is independent of E )
i  nAvd            E
m
11/22/2011                       Physics 122B - Lecture 9                             16
Example: The Electron Current
in a Copper Wire
The mean time between collisions for electrons in room-temperature
copper is  = 2.5x10-14 s.

What is the electron current in 2.0 mm diameter copper wire where
the internal field strength is E = 0.010 N/C?

e    (1.60 10-19 C)(2.5 10 14 s)(0.010 N/C)
vd     E                                            4.4  105 m/s
m                  (9.1110-31 kg)

n  8.5 1028 m-3 ; A   r 2   (1.0 103 m) 2  3.14 10-6 m 2

i  nAvd  1.2 1019 electrons/s
11/22/2011                   Physics 122B - Lecture 9                    17
Batteries and Current

11/22/2011         Physics 122B - Lecture 9   18
Electrical Current

 dQ                         
I     , in the direction of E 
 dt                         
1 ampere  1 A
 1 coulomb per second
= 1 C/s
DQ  I Dt

DQ eNe
I       ei
Dt Dt
11/22/2011             Physics 122B - Lecture 9   19
Example: The Current in a
Copper Wire

From the previous example,         i  nAvd  1.2 1019 electrons/s

What is the current I?

I  ei  (1.60 10-19 C)(1.2 1019 s -1 )  1.9 A

How much charge passes through the wire in an hour?

D Q  I Dt  (1.9 A)(3600 s)  6840 C

11/22/2011                    Physics 122B - Lecture 9                     20
Current and Electrons

11/22/2011         Physics 122B - Lecture 9   21
The Current Density in a Wire
I  ei  nevd A

I
J  current density   nevd
A

Example: A current of 1.0 A passes through a 1.0 mm diameter
aluminum wire. What is the drift speed of the electrons in the wire?

I  I        (1.0 A)
J  2                     1.3 106 A/m 2
A r    (5.0 104 m) 2
J         (1.3 106 A/m 2 )
vd                                       1.3  104 m/s
ne (6.0 1028 m -3 )(1.60 10 19 C)
11/22/2011                  Physics 122B - Lecture 9                   22
Kirchhoff’s Junction Law

 I  I
in           out

Ii
i    0; summed over all the currents to any "junction".

11/22/2011                  Physics 122B - Lecture 9            23
End of Lecture 9

 Before the next lecture, read Knight: 29.1 to 29.4.

11/22/2011            Physics 122B - Lecture 9          24

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